# Does the property of being "virtual" for a particle depend on the observer?

I've read at several places that a static magnetic (and electric for that matter) field can be thought of as made by virtual photons, at least that's what I understood.

Now, in Special Relativity we learn that an observer (in an inertial frame of reference) who sees only either an electric or a magnetic field will see both an $E$ and $B$ field if he starts moving. That is, he will see an electromagnetic field. And as far as I know an electromagnetic field is made of photons (not virtual photons).

Hence my question "is the property of being "virtual" for a particle (a photon in particular but any other particle is also welcome) dependent on the observer?"

• You might also want to look at this: physics.stackexchange.com/q/221842 Jan 17, 2016 at 3:09
• One talk about virtual photons in electric and magnetic field because the nature of these fields is not discovered. But somehow it is clear that photons are involved in electric and magnetic fields and in electromagnetic radiation too. Otherwise the (virtual) photons, which are inside the magnetic field, are not extractable, so how they give there properties to the EM radiation. Want to read more, see independent.academia.edu/HolgerFiedler and read my light-minded overview "Are photons composed particles" or the elaboration about "Complex one-dimensional structures of space" Jan 17, 2016 at 6:15

"Virtual" is not a property of a particle at all. And it is not true that pure electric and magnetic fields are "made out of virtual photons" or that a combined electromagnetic field is "made out of real photons".

A "virtual particle" is not a particle. It is an internal line in a Feynman diagram, which is in turn a graphic notation for a certain integral. No particle states are associated to such internal lines in the formalism. You may tell stories about them, but in the end, whatever they are, they are not particles in any sense.

From a field theory point of view, all static fields, whether electric, magnetic, the weak nuclear force or the strong nuclear force can be thought of as being mediated by virtual particles. [emphasis mine]

where now "mediated" does not mean that there are actual particles that "make up" the field, it means that the Feynman diagram between two charges in which a virtual photon is present gives rise to the electromagnetic force in the classical limit, see this question.

The relationship between the electromagnetic field (and with that also pure electric and magnetic fields), is complicated. The energy of the electromagnetic field corresponds to the number of photons in it, see this question, but this is a rather useless statement if the field state is not a number eigenstate. It is electromagnetic radiation, i.e. electromagnetic waves, which corresponds best to "being made out of photons", although also there the link is not clear cut, see this question.

• I almost downvoted this for saying that a virtual particle is an internal line in a Feynman diagram. I really wish you guys would point out the simplest case in which a particular theoretic construct shows up instead of always jumping to QFT. Virtual particles show up in vanilla quantum mechanics. Jan 17, 2016 at 3:07
• @DanielSank: What is your simplest example? Jan 17, 2016 at 3:09
• Second order perturbation theory. The virtual transitions are clear as day: $\sum_{\text{virtual state }m} \langle \text{final} | m \rangle \langle m | \text{initial}\rangle / (\text{some denominator, whatever})$. Jan 17, 2016 at 3:29
• So out of what the electric and the magnetic field and the EM radiation is made? Jan 17, 2016 at 6:19
• @ACuriousMind I think it would be useful to explain new ideas in the simplest possible context whenever possible. Here is an example where someone asked about contour shifting in the context of relativity but I explained it using a damped harmonic oscillator. Should you write an answer that long? Obviously that's up to you, but again I do think keeping it simple is helpful and clarifying for the reader. As I said, I just ask that we not always jump to relativistic QFT to explain simple things. Jan 17, 2016 at 18:20

The answer is actually simple and definitive. No.

Indeed they have a mass (different from the respective "real" particle), but we know that $P^2=-m^2$, and since the square modulus of a 4-vector is observer independent, a virtual particle is observer independent. (Curved space effect aside, like Unruh effect)

Still, by definition you cannot observe it, so...